作业帮设f(x)在区间[a,b]上连续,证明∫上限a,下限b.f(x)dx=∫上限a,下限bf(a+b-x)dx.
来源：学生作业帮助网 编辑：作业帮 时间：2024/05/05 05:25:03
![设f(x)在区间[a,b]上连续,证明∫上限a,下限b.f(x)dx=∫上限a,下限bf(a+b-x)dx.](/uploads/image/z/12678238-46-8.jpg?t=%E8%AE%BEf%28x%29%E5%9C%A8%E5%8C%BA%E9%97%B4%5Ba%2Cb%5D%E4%B8%8A%E8%BF%9E%E7%BB%AD%2C%E8%AF%81%E6%98%8E%E2%88%AB%E4%B8%8A%E9%99%90a%2C%E4%B8%8B%E9%99%90b.f%28x%29dx%3D%E2%88%AB%E4%B8%8A%E9%99%90a%2C%E4%B8%8B%E9%99%90bf%28a%2Bb-x%29dx.)
设f(x)在区间[a,b]上连续,证明∫上限a,下限b.f(x)dx=∫上限a,下限bf(a+b-x)dx.
设f(x)在区间[a,b]上连续,证明∫上限a,下限b.f(x)dx=∫上限a,下限bf(a+b-x)dx.
设f(x)在区间[a,b]上连续,证明∫上限a,下限b.f(x)dx=∫上限a,下限bf(a+b-x)dx.